# Core 3 (C3)

- You can find the formula booklet here and past papers from 2005 to present here. To convert your scores to UMS marks, my online converter tool and Android app are linked from the homepage of jameswalker.net.

As you can see in the formula booklet, nearly every formula and result on this page needs to be memorised.

You

**do not**need to know the numerical methods (decimal search and fixed-point methods) from the coursework for this exam. See the NM module for notes on these methods.

# Proof

**p1, p2** Types of proof:

**Proof by direct argument**uses a known fact - e.g. that $n^2$ is even if $n$ is even- With
**proof by exhaustion**, you test every possibility. For example, you could prove that there is only one prime number which is 1 less than a square number under 100 by subtracting 1 from all 9 squares under 100 and showing that only one of them is prime **Proof by contradiction**: for example, if you are trying to prove that $\sqrt{2}$ is irrational, you would show that it is false that $\sqrt{2}$ is rational. Therefore, if it's not rational, we know that it's irrational- With
**disproof by counterexample**, you can show that something is false by giving an example where the conjecture (thing you're trying to prove) fails. For example, you could disprove that*all odd numbers under 10 are prime*by showing that $9 < 10$ and 9 is divisible by 3

# Exponentials and natural logarithms

**a1, a2** $e^x$:

- $e$ is equal to 2.718 to 4 significant figures (it's irrational)
- On a graph of $y = e^x$, the gradient is equal to 1 at (0, 1)
- $e^x$ is its own derivative; $\frac{\d}{\dx}(e^x) = e^x$

## $\ln(x)$:

- $\ln(x)$ is shorthand for $\log_e(x)$
- The logarithm laws still apply for $\ln$, e.g. $\ln(x) + \ln(y) = \ln(xy)$ etc

**a3** Graphs:

- The graphs of $e^x$ and $\ln(x)$ are shown below. The blue is $y = e^x$ and red is $y = \ln(x)$
- The two graphs are inverse functions of each other - they are reflections of each other in $y = x$

- This means that $\ln(e^x) = x$, and similarly, $e^{\ln(x)} = x$

**a4** Exponential growth and decay:

- Usually the form $y = a \times b^x$:

- If $b > 1$, it's exponential growth

- If $b < 1$, it's exponential decay

# Functions

**f1** Mapping:

**Functions**map one or more inputs to outputs- Functions can either be
**one-to-one**(one input, which outputs one value), or**many-to-one**. This is because a function cannot produce two values - its purpose is to process a number into another. Therefore, a circle is not a function as any $x$ coordinate will map to two $y$ coordinates - There are
**one-to-many**and**many-to-many**mappings, but**these are not functions** - The
**input**is also called the**object** - The
**output**is also called the**image**

## Domain and range:

- The set of inputs is the
**domain**(e.g. the $x$-coordinates) - The set of outputs for the specific inputs used is the
**range**(e.g. the $y$-coordinates) - To state the type of numbers possible in a domain or range, use the $\in$ symbol followed by the symbol for the type. These are:

- $\Bbb R$: real numbers (any number that isn't the square root of a negative)

- $\Bbb Z$: integers

- $\Bbb Q$: rational numbers

- $\Bbb N$: positive integers (excluding 0) - For example, the domain of a function could be written as $x \in \Bbb Z, x \geq 3$, which tells us that $x$ must be an integer greater than 3
- Make sure to include this when defining a function - it is not called a function if part of the domain doesn't map to anything

## The codomain:

- Imagine the function $f(x) = x^2$ where $x \in \Bbb Z, x \neq 0$
- Because the domain only contains integers ($\Bbb Z$), it will therefore only map to positive square numbers
- We would write the range as $\{1, 4, 9, 16, 25, ...\}$
- The
**codomain**is the set of values that the range belongs to. Because the range is only positive integers, we would write the codomain as $y \in \Bbb N$

**f1, f8** Odd, even and periodic functions:

- A function is
**even**if it is symmetrical about the $y$-axis; $f(-x) = f(x)$ if $f(x)$ is even - A function is
**odd**if it looks the same when rotated 180° about $(0, 0)$; $-f(x) = f(-x)$ if $f(x)$ is odd **Periodic**functions repeat periodically (e.g. $y = \sin(x)$ repeats every 360°)- Functions can be more than one of the above, and some are none at all

**f2, f3, f4** Transformations:

- See C2 for how to apply transformations to $f(x)$
- If you need to do multiple transformations (e.g. $4f(x - 2)$), apply one at a time - with this example you could translate it to the right 2 places, and then stretch it ×4 in the $y$-axis

**f5** Composite functions:

- A
**composite function**is a combination of more than one function applied in series, e.g. $fg(x)$ is shorthand for the composite function $f$[$g(x)$] - Example question: $f(x) = x + 1$. $g(x) = x^2$.
*Find $fg(x)$*:**1)**Substitute $g(x)$ into $f$: $fg(x) = f(x^2)$**2)**Now do the same for $f$: $fg(x) = x^2 + 1$

**f6** Inverse functions:

- The inverse of $f(x)$ takes an output of $f(x)$ and finds the value of $x$ for it
- To find an inverse:

- Write in $y$ = _ _ _ form

- Swap $x$ and $y$

- Solve for $y$

- Replace $y$ with $f^{-1}(x)$ - A function $f(x)$ and its inverse $f^{-1}(x)$ are reflections in the line $y = x$
- An inverse function must only map to one output, as with a standard function

**f7** arcsin, arccos and arctan:

- Arcsin is often shortened to sin
^{-1}, and the same applies to arccos and arctan - The graphs of these functions is the graph of sin, cos or tan reflected in the line $y = x$ (they're inverse functions)
- For this, the domain must be restricted to create a one-to-one mapping:

- $y = \sin x$: domain = -90° ≤ $x$ ≤ 90°

- $y = \cos x$: domain = 0° ≤ $x$ ≤ 180°

- $y = \tan x$: domain = -90° < $x$ < 90°

**f9** The modulus function:

- Modulus ($|n|$) makes all values inside it positive; i.e. $|2| = 2$, $|-2| = 2$
- Therefore a graph of $y = |x|$ never goes below the x-axis, instead reflecting in the line $x = 0$

**f10** Solving equations with the modulus function:

- Change the modulus sign to brackets
- Now solve for $x$ to find the first solution
- To find the second solution, add a - sign in front of the brackets and solve again
- It often helps to sketch a graph
- The
*Abs*button on most calculators will input the modulus sign, this is useful for checking answers

# Calculus

**c1** The product rule (for u × v):

- When $y = u \times v$, $\frac{\dy}{\dx} = u$ $\frac{\d v}{\dx} + v \frac{\d u}{\dx}$
- Example: differentiate $(x^3+7x-1)(5x+2)$
**1)**Define $u$ and $v$: $u = x^3 + 7x - 1$, $v = 5x+2$**2)**$u \times \frac{\d v}{\dx} = 5(x^3+7x-1)$**3)**$v \times \frac{\d u}{\dx} = (5x+2)(3x^2+7)$**4)**Multiply the two and simplify to $20x^3 + 6x^2 + 70x + 9$ - $\frac{\d v}{\d x}$ is often shortened to $v'$, and $\frac{\d u}{\d x}$ to $u'$

**c2** The quotient rule (for u/v):

- For differentiating a fraction $\frac{u}{v}$:

$\frac{\dy}{\dx} = \frac{v\frac{\d u}{\dx} - u\frac{\d v}{\dx}}{v^2}$ *Order is important*; the numerator of the derivative must be $vu'-uv'$ and**not**$uv'-vu'$

**c3** The chain rule (for composite functions):

- $\frac{\dy}{\dx} = \frac{\dy}{\d u} \times \frac{\d u}{\dx}$
- Example: differentiate $y = (3x + 1)^2$
**1)**Differentiate the^{2}, leaving the inside of the fraction for now:

$\frac{\dy}{\d u} = 2(3x + 1)^{2-1} = 2(3x+1)$**2)**Differentiate inside the brackets: $\frac{\d u}{\dx} = 3$**3)**Multiply the two: $\frac{\dy}{\dx} = 3 \times 2(3x + 1)$**4)**Simplify: $\frac{\dy}{\dx} = 6(3x + 1)$ - After some practise this can be done by inspection - multiply the power by the inside of the bracket's derivative, then reduce the bracket's power by 1

**c4** Rate of change using the chain rule:

- $\frac{\d a}{\d b}$ is the rate of change of
*a*with respect to*b*, how much*a*changes if*b*changes by one unit - $\frac{\d a}{\d b}$ can always be expressed as a product of two fractions; for example $\frac{\d a}{\d c}$ × $\frac{\d c}{\d b}$. Therefore, the chain rule can be used

## An example problem:

*If a balloon is inflated at a rate of $10\ \mathrm{cm}^{3}$ per second, how quickly will the radius increase? (assume that the balloon is spherical)***1)**Write what we know: $\frac{\d V}{\d t} = 10$ where $V$ is the volume in $\mathrm{cm}^{3}$ and $t$ is the time in seconds**2)**Find an equation connecting two variables in the question: $V = \frac{4}{3} \pi r^3$ (volume of a sphere)**3)**Differentiate this: $\frac{\d V}{\d r} = 4\pi r^2$**4)**Write what we need to find out: $\frac{\d r}{\d t}$ (how quickly the radius increases with time)**5)**We now have $\frac{\d V}{\d t}$ and $\frac{\d V}{\d r}$. Find an equation: $\frac{\d V}{\d t} = \frac{\d V}{\d r} \times \frac{\d r}{\d t}$**6)**Substitute in what we know: $10 = 4 \pi r^2 \times \frac{\d r}{\d t}$**7)**Now rearrange: $\frac{\d r}{\d t} = \frac{10}{4} \pi r^2 = \frac{5}{2} \pi r^2$ $\mathrm{cm}\ \mathrm{s}^{-1}$

**c5** Differentiating an inverse function:

- If you know the derivative of a function $f(x)$, the derivative of $f^{-1}(x)$ is $\frac{1}{f'(x)}$
- Similarly, 1 over the derivative of the inverse will give the derivative of the original function

**c6** Implicit differentiation:

- This allows us to differentiate a function without $y$ being the subject
- The method uses the chain rule: $\frac{\d f(y)}{\dx} = \frac{\d f(y)}{\dy} \times \frac{\dy}{\dx}$
- Therefore, you just need to differentiate each $y$ part with respect to $y$ and multiply this by $\frac{dy}{dx}$

## Example:

*Differentiate $x^2 + y^2 + xy = 6$ with respect to $x$:*

$2x + 2y\frac{\dy}{\dx} + y + x\frac{\dy}{\dx} = 0$

Now solve for $\frac{\dy}{\dx}$:

$2y\frac{\dy}{\dx} + x\frac{\dy}{\dx} = -2x - y$

$\frac{\dy}{\dx} \times (2y + x) = -2x - y$

$\frac{\dy}{\dx} = \frac{-2x - y}{2y + x} = -\frac{2x + y}{x + 2y}$

**c7** Differentiating natural logs:

- $\frac{\d}{\dx}(e^x) = e^x$
- $\frac{\d}{\dx}(e^{ax}) = ae^{ax}$ (you could easily use chain rule instead)
- $\frac{\d}{\dx}(\ln(x)) = \frac{1}{x}$
- $\frac{\d}{\dx}(\ln(ax)) = \frac{1}{x}$ (because $\ln(ax) = \ln(a) + \ln(x)$ and $\ln(a)$ is a constant term which doesn't affect the gradient)

**c8** Differentiating trig functions:

- $\frac{\d}{\dx}(\sin x) = \cos x$
- $\frac{\d}{\dx}(\cos x) = -\sin x$
- $\frac{\d}{\dx}(\tan x) = \frac{1}{\cos^2 x} = \sec^2 x$

**c9** Integration by substitution:

- Example:
*find $\int e^{-4x}\ \dx$* **1)**Pick a substitution for $u$. This is usually given in the exam. Here we'll use $u = -4x$**2)**Now differentiate the expression for $u$: $\frac{\d u}{\dx} = -4$**3)**Rearrange this to make $\dx$ the subject: $\dx = \frac{\d u}{-4} = \frac{-1}{4}\ \d u$**4)**Now write the integral again with the substitutions: $\int e^{-4x}\ \dx = \int e^u \frac{-1}{4}\ \d u$**5)**This can now be integrated: $\int e^{-4x}\ \dx = -\frac{1}{4}e^u + c = -\frac{1}{4}e^{-4x} + c$- If there is more than one $x$, you need to get them all changed to $u$ (e.g. by rearranging the equation for $u$ to make $x$ the subject and then substituting this in)
- If it is a definite integral (with limits at the top and bottom of the $\int$ symbol), the limits also need to be substituted into the equation for $u$
- If you have a fraction, it should be easiest if you make $u$ the denominator. If not, it's usually best as the most complex term
- Remember that you can take constants outside of integrals. For example, $\int \frac{1}{4x}\ \dx = \frac{1}{4} \times \int \frac{1}{x}\ \dx$

**c10** Integrating $\frac{1}{x}$:

- $\int \frac{1}{x}\ \dx = \ln|x| + c$, i.e. the natural log of the modulus of $x$ plus $c$

**c11, c12, c13** Other integration results you need to know:

- $\int e^{ax}\ \dx = \frac{1}{a}e^{ax} + c$
- $\int \sin ax\ \dx = -\frac{1}{a}\cos ax + c$
- $\int \cos ax\ \dx = \frac{1}{a}\sin ax + c$
- $\int \frac{f'(x)}{f(x)}\ \dx = \ln|f(x)| + c$

**c14, c15** Integration by parts:

- This method can be used when you are integrating two things multiplied together (e.g. $\int xe^x\ \dx$). The method is shown below:
**1)**Choose $u$ and $v'$. $u$ should be the simplest part, or $\ln x$ if it contains a natural log. $v'$ should be the other**2)**Now find $u'$ by differentiating $u$**3)**Find $v$ by integrating $v'$**4)**Use the integration by parts formula $\int u\frac{\d v}{\d x}\ \dx = uv - \int v\frac{\d u}{\dx}\ \dx$:

- The left hand side is the integral you're finding

- The first term on the right is your values for $u$ and $v$ multipled together

- The last term on the right is your values for $v$ and $u'$ multiplied together, and then integrated**5)**Make sure that you remember to integrate the last part, it's easy to forget. Also remember $+ c$ if it's an indefinite integral